在php数据库中检索从php文件到swift的结果

时间:2016-02-06 20:12:46

标签: php mysql swift

我无法从php文件中获取数据。 我正在制作像myfitnesspal这样的健康和健身类型的应用程序。 我无法在php文件中输出查询的reuslt并在swift中打印结果。

这是php(getDetails.php)

<?php

require("Conn.php");
require("MySQLDao.php");
$email = htmlentities($_POST["email"]);

$returnValue = array();


$dao = new MySQLDao();
$dao->openConnection();
$userDetails = $dao->getUserDetails($email);

$returnValue["user"] =  $userDetails;

echo json_encode($returnValue);


$dao->closeConnection();

?>    

MySQLDao.php

public function getUserDetails($email)
{ 
$returnValue = array();
$test = array();
$sql = "select calories from users where username='" . $email . "'"; 

$result = $this->conn->query($sql);
if ($result != null && (mysqli_num_rows($result) >= 1)) {
$row = $result->fetch_array(MYSQLI_ASSOC);
$returnValue = $row;

if (!empty($row)) {
}
}
return $returnValue;
}

最后是我的快速代码

   let myURL = NSURL(string: "http://localhost:8888/getDetails.php");
         let request = NSMutableURLRequest(URL:myURL!);
       request.HTTPMethod = "POST";


   let postString = "email=\(userEmail)";

   request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);

    let task = NSURLSession.sharedSession().dataTaskWithRequest(request){
        data, response, error in

        do{

            let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary
            if let parseJSON = json{
                let resultValue = parseJSON["user"] as! String!;
                print("results:\(resultValue)")





            }

        }
        catch {print(error)}

    }

    task.resume();

0 个答案:

没有答案